1. General Solution – Homogeneous and Non-Homogeneous Equations
MATH 374 Lecture 12
General Solution – Homogeneous and Non-Homogeneous Equations

2. 4.5: General Solution of a Homogeneous Equation
The next theorem shows that if we know n solutions to an nth order linear, homogeneous equation, we know “all” of the solutions.
One key to this theorem is linear independence!
Theorem 4.4: Let {y1, y2, … , yn } be a linearly independent set of solutions of the homogeneous linear equation
b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = 0 (1)
for x 2 [a, b].
Suppose further that (1) is normal on [a, b].
If  is any solution of (1), valid on [a, b], then there exists constants ĉ1, ĉ2, … , ĉn such that
 = ĉ1 y1 + ĉ2 y2 + … + ĉn yn. (2) 2

3. Proof of Theorem 4.4 (Case when n = 2, other cases are similar.)
Let y1 and y2 be linearly independent solutions of
b0(x) y’’ + b1(x) y’ + b2(x) y = 0 (3)
on [a,b] and  be any solution of (3) on [a,b].
By Theorem 4.3, the Wronskian of y1 and y2 is non-zero at some x0 2 [a,b], i.e.
W(x0) = y1(x0) y2’(x0) –y1’(x0) y2(x0)  0. 3

4. Proof of Theorem 4.4 (continued)
W(x0) = y1(x0) y2’(x0) –y1’(x0) y2(x0)  0.
Proof of Theorem 4.4 (continued)
It follows from Theorem 4 in the Coddington Handout that the system
c1 y1(x0) + c2 y2 (x0) = (x0) (4a)
c1 y1’(x0) + c2 y2’(x0) = ’(x0) (4b)
has a unique solution, say c1 = ĉ1 and c2 = ĉ2.
Define the function f by:
f := ĉ1 y1 + ĉ2 y2. (5) 4

5. Proof of Theorem 4.4 (continued)
f := ĉ1 y1 + ĉ2 y2 (5)
Proof of Theorem 4.4 (continued)
From Theorem 4.1 (Principle of Superposition), it follows that f is a solution of (3) on [a,b], since f is a linear combination of solutions of (3) on [a,b].
From (5), (4a), and (4b), we see that
f(x0) = ĉ1 y1(x0) + ĉ2 y2 (x0) = (x0)
f’(x0) = c1 y1’(x0) + c2 y2’(x0) = ’(x0).
It follows from Theorem 4.2 (Existence and Uniqueness) that  = f = ĉ1 y1 + ĉ2 y2 on [a,b]. 
Note: We needed (1) to be normal to apply Theorems 4.2 and 4.3 in this proof! 5

6. General Solution; Fundamental Set
Definition: The general solution of the nth order linear homogeneous differential equation (1) is
y = c1 y1 + c2 y2 + … + cn yn
where {y1, y2, … , yn } are linearly independent solutions of (1) and c1, c2, … , cn are arbitrary constants.
We call {y1, y2, … , yn } a fundamental set of solutions of (1). 6

7. 4.6: General Solution of a Non-Homogeneous Equation
Theorem 4.5: Let yc be a solution of the homogeneous linear differential equation
b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = (1)
on a  x b and
let yp be a solution of the related non-homogeneous linear differential equation
b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = R(x) (2)
on a  x b.
Then y = yc + yp is also a solution of (2) on on a  x b. 7

8. b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = 0 (1)
b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = R(x) (2)
Proof of Theorem 4.5
Substituting y = yc + yp into the LHS of (2), we get:
b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y
= (b0(x) yc(n) + b1(x) yc(n-1) + … + bn-1(x) yc’ + bn(x) yc)
+ (b0(x) yp(n) + b1(x) yp(n-1) + … + bn-1(x) y’p + bn(x) yp)
= 0 + R(x) (since yc solves (1) and yp solves (2))
= R(x).  8

9. General Solution of a Non-Homogeneous Equation
b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = (1)
b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = R(x) (2)
General Solution of a Non-Homogeneous Equation
Theorem 4.6: Let {y1, y2, … , yn } be a linearly independent set of solutions of (1) on a  x  b and yp a particular solution of (2) on a  x  b.
Suppose that (2) is normal and Y is any solution of (2) on a  x  b.
Then there exists constants ĉ1, ĉ2, … , ĉn such that
Y = ĉ1 y1 + ĉ2 y2 + … + ĉn yn + yp. 9

10. Proof of Theorem 4.6 10 Consider the function Y – yp.
b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = R(x) (2)
Proof of Theorem 4.6
Consider the function Y – yp.
Substitute Y – yp into LHS of (2):
b0(x) (Y – yp)(n) + … + bn-1(x) (Y – yp)’ + bn(x) (Y – yp)
= (b0(x) Y(n) + … + bn-1(x) Y’ + bn(x) Y)
– (b0(x) yp(n) + … + bn-1(x) y’p + bn(x) yp)
= R(x) – R(x) (since Y and yp solve (2))
= 0.
By Theorem 4.4, there exist constants ĉ1, ĉ2, … , ĉn such that Y – yp = ĉ1 y1 + ĉ2 y2 + … + ĉn yn.  10

11. General Solution; Complementary Function
Definition: The general solution of the non-homogeneous linear differential equation (2) is
y = c1 y1 + c2 y2 + … + cn yn + yp
where {y1, y2, … , yn } are linearly independent solutions of (1), c1, c2, … , cn are arbitrary constants, and yp is any particular solution of (2).
We call yc = c1 y1 + c2 y2 + … + cn yn the complementary function of (2). 11

12. Example 1 12 Since yp = -11/12 -1/2 x is a particular solution of
y’’’ – 6 y’’ + 11 y’ -6 y = 3x, (3)
and it turns out (as we will find later) that the general solution to the homogeneous equation
y’’’ – 6 y’’ + 11 y’ -6 y = 0,
is yc = c1 ex + c2 e2x + c3 e3x,
it follows that the general solution to (3) is:
y = yc + yp = c1 ex + c2 e2x + c3 e3x -11/12 - 1/2 x. 12